Difference between revisions of "2006 AMC 12A Problems/Problem 17"
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s\left(s+r\sqrt{2}\right) \ | s\left(s+r\sqrt{2}\right) \ | ||
&= s^2+rs\sqrt{2}.\end{align*}</cmath> As in the first solution, we conclude that <math>r/s=\boxed{5/9}</math>. | &= s^2+rs\sqrt{2}.\end{align*}</cmath> As in the first solution, we conclude that <math>r/s=\boxed{5/9}</math>. | ||
+ | |||
+ | ===Solution 5 - Answer Choices=== | ||
+ | |||
+ | We roughly measure the distance of <math>r</math> and the distance of <math>y</math>. Since <math>r</math> is clearly less than <math>s</math>, we can eliminate answer choices (D) and (E). Next, if we compare the distances, <math>r</math> seems to be just a little more than half of <math>s</math>, thus eliminating answer choice (A). <math>r</math> is only a little bit bigger than half of <math>s</math>, so we can reasonably assume that their ratio is less than <math>\frac{3}{5}</math>. That leaves us with answer choice <math>\boxed{C}</math> , or <math>\frac{5}{9}</math>. | ||
== See Also == | == See Also == |
Revision as of 23:13, 26 October 2022
Contents
[hide]Problem
Square has side length
, a circle centered at
has radius
, and
and
are both rational. The circle passes through
, and
lies on
. Point
lies on the circle, on the same side of
as
. Segment
is tangent to the circle, and
. What is
?

Solutions
Solution 1
One possibility is to use the coordinate plane, setting at the origin. Point
will be
and
will be
since
, and
are collinear and contain a diagonal of
. The Pythagorean theorem results in
This implies that and
; dividing gives us
.
Solution 2
First note that angle is right since
is tangent to the circle. Using the Pythagorean Theorem on
, then, we see
But it can also be seen that . Therefore, since
lies on
,
. Using the Law of Cosines on
, we see
Thus, since and
are rational,
and
. So
,
, and
.
Solution 3
(Similar to Solution 1)
First, draw line AE and mark a point Z that is equidistant from E and D so that and that line
includes point D. Since DE is equal to the radius
,
Note that triangles and
share the same hypotenuse
, meaning that
Plugging in our values we have:
By logic
and
Therefore,
Solution 4 - Alcumus
Let ,
,
,
, and
. Apply the Pythagorean Theorem to
to obtain
from which
. Because
and
are rational, it follows that
and
, so
.
OR
Extend past
to meet the circle at
. Because
is collinear with
and
,
Also,
which implies
, so
is an isosceles right triangle. Thus
. By the Power of a Point Theorem,
As in the first solution, we conclude that
.
Solution 5 - Answer Choices
We roughly measure the distance of and the distance of
. Since
is clearly less than
, we can eliminate answer choices (D) and (E). Next, if we compare the distances,
seems to be just a little more than half of
, thus eliminating answer choice (A).
is only a little bit bigger than half of
, so we can reasonably assume that their ratio is less than
. That leaves us with answer choice
, or
.
See Also
2006 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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